Problem: Solve for $x$ and $y$ by deriving an expression for $x$ from the second equation, and substituting it back into the first equation. $\begin{align*}9x-6y &= 3 \\ 6x+7y &= -5\end{align*}$
Begin by moving the $y$ -term in the second equation to the right side of the equation. $6x = -7y-5$ Divide both sides by $6$ to isolate $x$ $x = {-\dfrac{7}{6}y - \dfrac{5}{6}}$ Substitute this expression for $x$ in the first equation. $9({-\dfrac{7}{6}y - \dfrac{5}{6}}) - 6y = 3$ $-\dfrac{21}{2}y - \dfrac{15}{2} - 6y = 3$ Simplify by combining terms, then solve for $y$ $-\dfrac{33}{2}y - \dfrac{15}{2} = 3$ $-\dfrac{33}{2}y = \dfrac{21}{2}$ $y = -\dfrac{7}{11}$ Substitute $-\dfrac{7}{11}$ for $y$ in the top equation. $9x-6( -\dfrac{7}{11}) = 3$ $9x+\dfrac{42}{11} = 3$ $9x = -\dfrac{9}{11}$ $x = -\dfrac{1}{11}$ The solution is $\enspace x = -\dfrac{1}{11}, \enspace y = -\dfrac{7}{11}$.